Original solution: 50 mL of 0.05 M H2PO4 moles of H2PO4 = Concentration x Volume = 0.05M X 50 X 10-3 L = 0.0025 moles Na

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Original Solution 50 Ml Of 0 05 M H2po4 Moles Of H2po4 Concentration X Volume 0 05m X 50 X 10 3 L 0 0025 Moles Na 1 (114 KiB) Viewed 37 times

In a lab, the pH was 11. Does the calculated "predicted" pH
match the pH from the lab?
Original solution: 50 mL of 0.05 M H2PO4 moles of H2PO4 = Concentration x Volume = 0.05M X 50 X 10-3 L = 0.0025 moles NaOH Added: 205 uL5M NaOH mmoles of NaOH = Concentration x Volume = 5M X 205 x 10-6L = 0.001025 moles ICE Table for the reaction of H2PO4 with NaOH H2PO4 + OH <----------> HPO 2 + H20 Initial 0.0025 mol 0.001025 mol Change -0.001025 - 0.001025 +0.001025 Equilibrium 0.001475 0 0.001025 = [H, PO4-) = = 0.001025 mol 0.001025 mol (HPO42-] = 0.020416 M 205x10-6 + 50x10-3 L 0.050205 L 0.001475 mol 0.001475 mol = 0.029379 M 205x10-6 + 50x10- L 0.050205L pH can be calculated using Hasselbach – Henderson equation (HPO ?,] [H, PO4 pH = 7.2 + log 0.020416 0.029379 pH = 7.2 - 0.1581 pH = 7.04 pH = pKx + log (